How do you use trigonometric substitution to write the algebraic expression #sqrt(x^2+25)# as a trigonometric function of #theta# where #0<theta<pi/2# and #x=5tantheta#?

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Answer 1 · 2017-06-24T12:58:45

#g(theta)=5sectheta#

Let #sqrt(x^2+25)#

= #sqrt((5tantheta)^2+25)#

= #sqrt(25tan^2theta+25)#

= #sqrt(25(tan^2theta+1))#

= #5sqrt(sec^2theta)#

= #5sectheta#

Hence #f(x)=sqrt(x^2+25)# can be written as #g(theta)=5sectheta#